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If `cosy=xcos(a+y),` with `cosa!=+-1,` prove that `(dy)/(dx)=(cos^2(a+y))/(sina)dot`

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Given relation is cos y = x cos (a+y). Therefore,
`x=(cos y)/(cos (a+y))`
Differentaiting w.r.t.y, we get
`(dx)/(dy)=(d)/(dy)((cos y)/(cos (a+y)))`
`=((cos (a+y)(-sin y)- cos y (-sin (a+y)))/(cos^(2)(a+y)))`
`=((-cos (a+y) sin y + cos y sin (a+y))/(cos^(2) (a+y)))`
`=((sin (a+y-y))/(cos^(2)(a+y)))=(sin a)/(cos^(2)(a+y))`
`therefore" "(dy)/(dx)=(cos^(2)(a+y))/(sin a)`

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